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Let $\newcommand{\ID}[1]{\langle#1\rangle}F$ be the quotient ring $\mathbb{Q}[x]/\ID{x^3}$, where $\mathbb{Q}$ is the field of rational numbers.

Then find out which are correct

(i) There are exactly three distinct proper ideals of $F$

(ii) There is only one prime ideal in $F$

(iii) $F$ is an Integral Domain

Answer:

The ideals of $F$ are $\ID{0}$, $\ID{x}$, $\ID{x^2}$ and $\ID{x^3}$

The proper Ideals are $\ID{x}$ and $\ID{x^2}$

Thus there are two distinct ideals in $F$

But the option (i) is given to be correct.

Further $\ID{x}$ is the only prime ideal of $F$.

Thus option (ii) should be correct.

But how option (i) is correct?

Why $\ID{x^3}$ is a proper ideal?

Help me out, please.

egreg
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MAS
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    What do you mean by $\mathbb{Q} / \langle x^3 \rangle$? Possibly $\mathbb{Q}[x] / \langle x^3 \rangle$? Note that it is not a quotient field (otherwise it would not make sense to ask for its ideals). – 57Jimmy Mar 31 '18 at 15:31
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    You may mean $\mathbb{Q}[x]/\langle x^3\rangle$! – Qurultay Mar 31 '18 at 15:33
  • yes , absolutely . it is my mistkae – MAS Mar 31 '18 at 15:41
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    Did you mean $\langle1\rangle$ (also known as $\langle x^0\rangle$ and as $\mathbb Q[x]/\langle x^3\rangle$) rather than $\langle0\rangle$ in your list of ideals? I ask because $\langle0\rangle=\langle x^3\rangle$, so you've listed this ideal twice. – Andreas Blass Mar 31 '18 at 16:15
  • But how $ \ <x^3> \ $ becomes a $ \ proper \ \ Ideal \ $ ? Kindly help me with this – MAS Mar 31 '18 at 16:18
  • @yourmath A proper ideal is any ideal which is not the whole ring. This includes the zero ideal, which corresponds to the ideal generated by $x^3$ – Mike Earnest Mar 31 '18 at 16:25

1 Answers1

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I wouldn't say that $\newcommand\ID[1]{\langle #1\rangle}\ID{x}$ is an ideal in $F$, although it's a common abuse of language.

Precisely, the ideals of $F$ are in one-to-one correspondence with the ideals of $\mathbb{Q}[x]$ that contain $\ID{x^3}$. Since this ring is a principal ideal domain, such ideals are $\ID{1}$, $\ID{x}$, $\ID{x^2}$ and $\ID{x^3}$.

If $I\supseteq\ID{x^3}$ is an ideal in $\mathbb{Q}[x]$, the corresponding ideal in $F$ is $I/\ID{x^3}$. So you have three proper ideals. A proper ideal is one that is not the whole ring and certainly $\ID{x^3}/\ID{x^3}$ (the zero ideal in $F$) is not the whole ring.

Option (ii) is also correct, because $\ID{x}$ is the only prime ideal in $\mathbb{Q}[x]$ containing $\ID{x^3}$.

Option (iii) is incorrect, because $\ID{x^3}$ is not a prime ideal.

egreg
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  • why $ \ <x^3> \ $ is a proper Ideal? – MAS Mar 31 '18 at 16:33
  • @yourmath It is actually the zero ideal in $F$ and it is obviously proper, that is, $\ne F$. – egreg Mar 31 '18 at 16:48
  • @egreg how does ideal of $F$ are in one-one corespondence with ideals of $\mathbb{Q[x]}$? – TheStudent Nov 19 '19 at 08:17
  • @TheStudent Not with all ideals of $\mathbb{Q}[x]$, but only with those that contain $\langle x^3\rangle$. This is one of the homomorphism theorems: if $f\colon R\to S$ is a surjective ring homomorphism, there is a bijection between the ideals of $S$ and the ideals of $R$ containing $\ker f$; the bijection is defined by $I\mapsto f^{-1}(I)$, for $I$ an ideal of $S$. – egreg Nov 19 '19 at 08:36